SLM_(1)

Matrices 71 The augumented matrix is: 3 2 1 9  1 1 1 9   3 1 1 1  (9) Singular and Non-singular Matrix: If the determinant of a matrix A is equal to zero, then it is said to be a ‘singular matrix’. If the determinant of A is not zero then the matrix A is said to be a ‘non-singular matrix’. Ÿ Symmetric Matrix: A matrix that is equal to its transpose is said to be symmetric.  2 3 4  2 3 4 3 7 8  3 7 8 For example if A =   then A' =  4 8 5  4 8 5    and A = A', therefore A is said to be symmetric. If a matrix A = – A', then it is said to be ‘skew-symmetric’.  07 4 3  0 7  For example  0 4 3

4.3 Addition of Matrices Two matrices A and B are said to be conformable for addition if both the matrices have the same number of rows and same number of columns. aa  11 12 a13  a a a  If A =  21 22 23  a a   31 32 33  bb b  11 12 13  b b b  and B =  21 22 23  b b b   31 32 33  CU IDOL SELF LEARNING MATERIAL (SLM)

72 Business Mathematics and Statistics aaa  11 + b11 12 + b12 13 + b13  a +b a +b a +b  then A + B =  21 21 22 22 23 23  a +b a +b a +b   31 31 32 32 33 33   a11 b11 ab a 13 b13  and A – B = a b 12 12  21 21 ?b ? 23 b23 22 22  ? 33 b33   ? 3  a31 b31 b 2 32 With reference to addition, the Commutative, Associative and Distributive laws hold good. aa bb cc  11 12   11 12   11 12  a aFor example, if A   and B =  b b  and C = c c   21 22   21 22   21 22  then (i) A+B=B+A (Commutative law) (ii) (A+B)+C=A+(B+C) (Associative law) (iii) K(A+B)=KA+KB (Distributive law) where K is a certain constant. 4.4 Multiplication of Matrices Two matrices A and B are siad to be conformable for multiplication if the number of columns of A is equal to the number of rows of B.

If A is an m × p matrix and B is p × n matrix, then AB will be an m × n matrix. Here it is not possible to get the product BA as the number of columns of B is not equal to the number of rows of A, i.e., n m.  2 1 4  1 2  3 5 Example: 12: Given A =  B=  3 2 1  1 4  21+13 41 22+15 44 AB=   31+23 11 32+25 14   9 25  ?   10 20  CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 73 4.5 Applications of Matrices In this section, applications of matrix addition and subtraction will be illustrated with suitable examples. Example 1: A firm produces two products A and B using three plants P1, P2 and P3. During the month of January, P1 produced 160 unit of A and 80 units of B; P2 produced 90 units of A and 120 units of B; and P3 produced 200 units of A and 170 units of B. During the month of February, P1 produced 140 units of A and 70 units of B; P2 produced 100 units of A and 110 units of B; and P3 produced 180 units of A and 150 units of C. Find the total production of these two products for two months. Solution: Let P be the matrix which represents the production in the month of January. A B §160 80 ·P ¨ ¸1 ? P=¨90 120¸P2 ¸ ¨ ©200 170¹P3 Let Q be matrix which represents the production in the month of February. 8. B §140 70 ·P ¨ ¸1 ? Q =¨100 110¸P2 ¨ ¸

©180 150¹P3 The total production (say R) is given by R = P + Q AB §160 80 · §140 70 · §300 150·P ¨ ¸¨ ¸¨ ¸1 90 120¸ 100 110¸ = 190 230 P ?R =¨ +¨ ¸¨ ¸2 ¨ ¸¨ 150¹ ¨ ¸ ©200 170¹ ©180 ©380 320¹P3 ? P1 produced 300 units of A and 150 units of B; P2 produced 190 units of A and 230 units of B; and P3 produced 380 units of A and 320 units of B. Applications of Scalar Multiplication Sometimes, we come across the situations in which a set of data is to be multiplied by a constant. This is illustrated by following example. CU IDOL SELF LEARNING MATERIAL (SLM)

74 Business Mathematics and Statistics Example 2: A firm has four plans P1, P2, P3 and P4 and produces two products X and Y. It is assumed that the daily production is uniform. Number of units produced per pay is as follows: PPPP 1234 X 100 80 90 110 Y 60 70 40 50 Find the output of each product from each plant during a month of 30 days. Solution: Let A be the matrix which represents the daily production. P1 P2 P3 P4 10. §100 80 90 110·X A=¨¸ ©60 70 40 50¹Y 13. The total production (say B) in 30 days is given by B = 30A P1 P2 P3 P4 §3,000 2, 400 2,700 3,300·X ? B=¨ ¸ 1,800 2,100 1,200 1,500 Y ©¹ ? This the required production by four plants. Applications of Matrix Multiplication Multiplication of matrices plays a major role in illustrating business applications. In this section we will see several examples of different business applications.

Example 3: A furniture manufacturing company manufactures 50 table and 40 chairs in a day. Each table requires 4 hours of labour and 5 units of material. A chair requires 2 hours of labour and 3 units of material. Using matrices, find how many labour hours and material will be required each day? Solution: Let A be the matrix which represents the number of tables and chairs. §50·T ¨¸ ? A = 40 C ©¹ 2 u1 Let B be the matrix which represents the input per unit C 14. 4 2·L ¸ ? B=¨ © 5 3¹M2 u2 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 75 Total input (say C) is given by C =BA TC §280·L C=L§4 2·§50·T =¨ ¸ M¨ 5 ¸ ©370¹M 3¸¨40 C © ¹© ¹ 15. 280 labour hours and 370 units of material are required. 4.6 Orthogonal Matrix a  1 b1 c1  a b c  If A =  2 2 2  , then a b c   3 3 3 It is said to be orthogonal if the sum of the squares of the elements of any rwo or of any column is unity and the sum of the products of the corresponding elements of any two rows or of any two columns is zero. 4.7 Inverse of a Matrices If A and B are two square matrices of the same order and if AB = BA = I, where I is the unit matrix then B is said to be the inverse of A and it is denoted by A–1.

4.8 Cofactor Matrix aa a  11 12 13  a a a  If A =  21 22 23   a a a   31 32 33 then the matrix formed by the cofactors. A11 A12 A13 A21 etc of the elements a11 a12 a13 a21 etc; respectively which is represented as CU IDOL SELF LEARNING MATERIAL (SLM)

76 Business Mathematics and Statistics A AA 11 12 13   A A A   21 22 23  A A A   31 32 33  is called the Cofactor matrix of A. 4.9 Adjoint Matrix The transpose of the Co-factor matrix is called the Adjoint Matrix. Calculation of the Inverse of a matrix A The inverse of A, denoted by A–1 is expressed as the Adjoint of matrix A divided by the determinant of A. Matrix A is non-singular. A–1 is also called ‘reciprocal of matrix A’. Thus A–1 = Adj A A 4.10 Method of Solving Simultaneous Equations Using Matrix Property

Suppose a1x + b1y + c1z = P1 a2x + b2y + c2z = P2 a3x + b3y + c3z = P3 Using the matrix notation we can write the above equations as  a 1 b1 c1  x   P1   a b c  y  =  P 2 2 2    2 b a3 c   z   P  3 3     3 i.e., AX = P 2 A–1 AX = A–1P i.e., IX = A–1P i.e., X = A–1P

CU IDOL SELF LEARNING MATERIAL (SLM) Matrices 77 3 2 6   8 2 5   569 361 Example 13: If A =   B=   Find (i) 2A–3B  2 7 3   743  (ii) AB+BA Solution:  6 24 4 6 12 15   12 18 183  199  2A–3B=    421 1412 6 9 30 2 3  1 6 15  =   25 2 3   60 18 1   9 7 45   82 8  49  121 19 39 + AB+BA=    26 50 26   7 59 3   64 25 46 

 131  140 47 =   19 109 29  3 4 1 2 1 3  2 1 4  2   1 5 3 24 42 3 Example 14: Given A =   B=  C=    1 14   5 1 3  1 6 2  Find the following (i) 2A–3B+C (ii) A' and B' (iii) C–1 Solution: 10 6 3   96 1 (i) 2A – 3B + C =    12 11 15 

CU IDOL SELF LEARNING MATERIAL (SLM)

78 Business Mathematics and Statistics 3 2 1 2 3 5 4 11 1  21 (ii) A' =   and B' =   1 3  5 4   4 3 14 22 5  Adj C 1  20 =  11 0 (iii) C' = C 121    26 11 16  Example 15: Find the inverses of A , B , C where 3 2 4   3 1 4 6    1 1 61 2  A=  B=  3  2 3 5 5  2 1 2 3  4 5 2 C=    2 6 -4  3 2 4   1 6 1 Solution: For matrix A =    2 3 5

(i) |A| = – 3  -6 1  1 1 1 -6 –2  –4   3 -5  2 -5  2 3 = –3(30–3)–2(–5–2)–4(3+12) = –3(27)–2(–7)–4(15) = –81+14–60 = –141+14 = – 127 (ii) The co-factor of Matrix A is AA A  11 12 13   A A A  21 22 23 A A A  31 32 33  A 6 1  11 =   = 30–3=27 3 5 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 79 A 1 1 12 = –   = –(–5–2)=7 2 5 1 6  A= 13   =3+12=15  2 3 A  2 4  21 = –  = –(–10+12)=–2  3 15 A 3 4 = 22   = 15+8=23  2 5 A 3 2 23 = –   = –(–9–4)=13  2 3  2 4  A31 =   =2–24=–22  6 1 3 4 A =–  = –(–3+4)=–1 32   11 3 2  A33 =   =18–12=16  1 6 Co-factor matrix

 27 7 15  2 23 13  A=    22 1 16  (iii) The transpose of the factor matrix of A is the Adjoint matrix A, where  27 2 22   7 23 1 Adj A =    15 13 16  CU IDOL SELF LEARNING MATERIAL (SLM)

80 Business Mathematics and Statistics (iv) Inverse of the matrix A is 22   27 2  127 127 127  23 1  Adj A  7   –1 =  127 127 127  A = |A| 15 13 16   127 127 127  To verify the above solution we should prove that  1 0 0 –1 –1 = I = 0 1  0 A A= AA  0 0 1   27 2 22   127 127 127   3 2 4  7 23 1 –1   1 6   1 ? A A =  127 127  127  15 13 16 2 3 5   127 127 127   81 2 44 52 12 66 108 2 110  ++ ++  127 127 127 127 127 127

127 127 127   21 23 2 14 138 3 28 23 5  + 127 +  30 =  127 127 127 127 127 127 127 127  45 13 32 78 48 60 13 80   127 127 127  127 127 127 127 127 127   127 0 0    1 0 0  127  127 127 127  0 127 0  0 1  0 = =   127 127 127  0  0 1 0 0 127   127 127  Thus it is verified. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 81 Example 16: For matrix B 3 1 4 6  1 =2    5 2 3 1 6 2 6 2 1   (i) |B| = 3   +1 +4 3 2 2 3 5 5      = 3(3–12)+1(–6–30)+4(4+5) = 3(–9)+1(–36)+4(9) = –27–36+36 = – 27 0 (ii) The co-factor Matrix of B is B  B11 12 B13   B B  B21 22 23 B B B   31 32 33  B 1 6 11 =  2  = 3–12=–9   6 B 2  = –(–6 –30)=36 12 = –  5  

B 2 1 13 =  5  =4+5=9  2 B 1 4 3 21 = –  2  =–(3–8)=5  ª3 4º B22= ¬«5 3¼» = –9 – 20 = –29 B 3 1 23 = –   = –(6 + 5)= –11 5 2  B 1 4 31 =   = –6+4=–2  1 6 CU IDOL SELF LEARNING MATERIAL (SLM)

82 Business Mathematics and Statistics 3 4 B = –  = –(18– 8) = –10 32  = –3+2=–1 2  6 B 3 1 33 = 2   ? Co-factor matrix 9 36 9 29 5  11 B=    2 10 1  (iii) The transpose of the cofactor matrix of B / Adj. B 9 5 2   36 29 10    9 11 1  ? The required inverse of the matix B is 9 5 2  B–1 = Adj B  27 27 27   36 29 10  B   = 27 27 27

9 11 1     27 27 27 To verify the above solution we should prove that B–1 B = B B–1 = I  9 5 2  27 27 27   36 29 10   3 1 4    1 6 –1  27  2 3 B B= 27 27   2 9 11 1  5   27 27 27  CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 83 27 10 10 9 5 4 36 30 6  + +  27 27 27 27 27 27 27 27   27 58 50 36 29 20 144 174 30   108 + +  27 27 27 27 27 27 27 27  = 27  25 9 11 2 36 66 3  27 + +  27 27 27 27 27 27 27 27 27   1 0 0  1  0 0 =   001 Thus it is verified. Example 17: For matrix C we have 1 2 3 4 5  2   C=  2 6 4  (i) |C|= 1(–20–12)–2(16+4)–3(24–10) = –32– 40–42 = – 114 0

(ii) By calculating different cofactors we obtain the following co-factor of matric C. CC  C11 12 13  C C C  C =  21 22 23  CC C  31 32 33   32 20 14   2  20 10 = 11 14   13   CU IDOL SELF LEARNING MATERIAL (SLM)

84 Business Mathematics and Statistics (iii) Adj. of C.  32 26 11  2  20 10 14 =   14 13   (iv) Inverse of Matrix C  32 26 11  Adj. C  114 114 114    20 2 14   C–1 = =  114 114 114   14 10 13  C   114 114 114  To verify C–1 C = C C–1 = I  32 26 11   114 114 114   1 2 3 4 5 2     20 2 14   

C–1C=  114 114 114    6 4   2 10 13  14    114 114   114  32 + 104 22 64 130 66 96 52 44   114 114 114 114 114 114 114 114 114   20 8 28 40 56  10 84 60 4 114 114  + 26 28 114 114 114 114 50 78 42 20 114 =  114 114 114 114 52  14 40 114 114 114 114  + 114    114 114 1 0 0  010 =  001  CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 85 Thus it is verified. 8 2 4 –1 Example 18: If A =   find A 3 5 5 Solution:  5 1 3  (i) |A|=8(15–5)–2(9–25)+4(3–25) = 8(10)–2(–16)+4(–22) = 80+32–88 = 112 – 88 = 24 0 (ii) Cofactor matrix of A is:  10 16 22  4  2 2 A=    10 28 34   10 2 10  4   (iii) Adj. A = 16 28    2 2 34 

 10 2 10  Adj. A  24 24 24   16 4 28   (iv) A–1 = = 24 24 24  22 2 34  A   24 24 24  (v) To verify we should prove that A–1A= AA–1 = I CU IDOL SELF LEARNING MATERIAL (SLM)

86 Business Mathematics and Statistics  10 2 10     8 2 4  24 24 24   4 28 16 355  –1   24 24  ?AA= 1 2 34   24  5 3 22    24 24 24   80 6 50 20 10 10 40 10 30   24 24 24 24 24 24 24 24 24   128 12 140 32 20 28 64 20 84  + + 24 +  =  24 24 24 24 24 24 24 24  176 6 170 44 10 34 88 10 102   ++ ++ ++    24 24 24 24 24 24 24 24 24  24 0 0   24 24 24  1 0 0    0 24 0   24 24 24 

= 010 = =I   0 0 1 0 0 24     24 24 24  ? It is proved.  1 4 2 2 5  –1 Example 19: Given A =  1  find A 7 5  3 Solution: (i) |A| = 1 (25 – 3) – 4 (10 – 6) + 2 (6 – 30) = 22 – 4 (4) + 2 (–24) = 22–16–48 = 22–64 = – 42 0 CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 87 (ii) Cofactor matrix of A =  22 4 24   14 7 21   6 3 3  22 14 6 4 7  3   (iii) Adj. A =  24 21 3  22 14 6   42 42 42  4 7 3 Adj. A  (iv) A–1 = =  42 42 42 A  24 21 3    42 42 42  (v) To verify we should prove that A–1A=AA–1 = I 22 14 6   42 42 42   4 7 3   1 4 2   A–1A=  42 42 42  2 5 1 

24 21 3    635   42 42 42  22 28 36 88 70 18 44 14 30 + +  ++ ++ 42 42 42   42 42 42 42 42 42 8 7 15   4 14 18 16 35 9 + + +  =  42 42 42 42 42 42 42 42 42  24 42+18 96 105 + 9 48 21 + 15    42 42 42 42 42 42 42 42 42   1 0 0  0 1 0 =  . Hence verified. 001 1 3 2 –1 Example 20: Given A = 2 4  5   find A 1 3  6 CU IDOL SELF LEARNING MATERIAL (SLM)

88 Business Mathematics and Statistics Solution: z |A|=(12–30)+ 3(6 + 5)+2(–12– 4) – 18 + 3 (11) + 2 (–16) –18+33–32 – 17 0 z Cofactor matrix of A 18 11 16 1 3  3  A=  3 10  7  18 3 7   1  11 9 (iii) Adj. A =   16 3 10  18 3 7   17 17 17  9  1 11 Adj. A  (iv) A–1 = = 17 17 17  A  16 3 10 

 16 17 17  (v) To verify we should prove that A–1 A = A A–1 = I  18 3 7   17 17 17  9  1 1 3 2 11     17  2 4 5 1  –1 17 17 10 63  A A=  16 3    17 17 17  18 6 7 54 12 42 36 15 21 + ++ 17   17 17 17 17 17 17 17 17  5  11 2 9 33 4 54 22 27   17 17 17 + +  = 17 6 10 48 17 17 17 17 17   16 12 60 32 15 30   ++  17 17 17 17  17 17 17 17 17  CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 89  1 0 0  0 1 0 z  =I 001 Hence verified. 4.10 Summary In the introduction, we study the meaning and definition of the Matrix and the various types of Matrices such as Null Matrix, Diagonal Matrix, Scalar Matrix, Unit Matrix, Equal Matrices, Transpose of a Matrix, Coefficient Matrix, Augmented Matrix, Singular and Non-singular Matrix, Symmetric Matrix. We learn the operation of Matrix Additiion and Multiplication of Matrices. We understand the meanings of Orthogonal Matrix, Co-factor Matrix, Adjoint Matrix and the method of calculation of the inverse of a matrix by standing the different stages through the illustrative examples. 4.12 Key Words/Abbreviations Matrics called: Null, Diagonal, Scalar, Unit, Transpose, Coefficient, Augmented, Singular, Non-singular, Symmetric, Orthogonal, Inverse, Co-factor, Adjoint. 4.13 Learning Activity

Learn the operations of Matrix Addition and Multiplication and the calculation of the Inverse of a Matrix by working out the examples given in the ‘Exercises on Matries’ ............................................................................................................................................................. ............................................................................................................................................................ 4.14 Unit End Questions (MCQ and Descriptive) A. Descriptive Type: Short Answer Type Questions [Ans: x = 1, y = –2] © Solve, using determinants 7x – 4y = 15 3x + 5y + 7 = 0 CU IDOL SELF LEARNING MATERIAL (SLM)

Business Mathematics and Statistics 90 2. Evaluate the determinant of the matrix.  a a+3 a + 6 [Ans: 0]   a+1 a+4 a+7    a + 2 a + 5 a + 8 3. Evaluate the determinant of the matrix  35 38 24  15  12 8   [Ans: 360]  51 57 40  7 1 1  2 1 4. Find the inverse of the matrix 10 , and hence solve the following sytem  equations:  6 3 2 7x – y – z = 0, 10x – 2y + z = 8, 6x + 3y – 2z = 7. 2 4 1  2 and hence solve the following system of 5. Obtain the inverse of the matrix 3 1   1 3 -3 equations. 2x + 4y – z = 9, 3x + y + 2z = 7, x + 3y – 3z = 4 ¡ 2 1¯ 6. If the matrix A is given by ¡ ° , prove that A satisfies the relation A2 – 4A + 3I = °

¢1 2 ± ª4º 0 where I stands for the unit matrix of order 2. «0» 7. Let A and B be two matrices defined by A = [1 –½ –1 3] and B = « » Obtain the «2» «» product AB and BA. Find, if possible, the sum of the two matrices AB and BA. 8. Let two matrices A and B be given by 1 1 0  2 2 4  2 3   1 4 A= 42 4  and B =  0 2  2 1 5    Verify that BA = 6I, where I is the unit matrix of order 3. CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 91 z A company is considering which of the three methods of production it should use in producing three goods A, B and C. The amount of each good produced by each method is shown in the matrix. (C.A.) ABC Method 1 4 8 2 Method 2 5 7 1 Method 3 5 3 9 The vector (or row matrix) (10 4 6) represents the profit per unit for the goods. A, B and C in that order. Using matrix multiplication, find which method maximises total profit. 10. An engineering product requires five kinds of materials, the quantities of which are given in the form of the vector (i.e., a row matrix) q = [17 4 36 18 6]. If p = [5 200 4 50 3] represents a vector of the corresponding prices per unit in Rs. find the total expenses for the manufacture of the product. 1 2 3  2  1 , show that A3 –23A – 40I = 0, where I is the unit matrix of order 2 1  3 11. If A =  4  3 and 0 is the null matrix of order 3.

7. There are two families A and B. Threre are 2 men, 3 women and 1 child in family A and 1 man, 1 woman and 2 children in family B. The recommended daily allowance for calories is: Men :2400, Women: 1900, Children: 1800 and for proteins is: Men: 55 gms, Woman: 45 gms, Children: 33 gms. Represents the above information by matrices. Using multiplication, calculate the total requirements of calories and proteins for each of the two families. 1 2 3  4 5 2 13. Find the inverse of the matrix.    3 5 6  CU IDOL SELF LEARNING MATERIAL (SLM)

92 Business Mathematics and Statistics 14. 2 ª 4 1 2º what is the Adjoint matrix of A, Where A = « «» 15. Solve the system of linear equations. 10 4»? x+y+z=6 «» x + 2y + 3z = 14 –x+y–z=–2 ¬ 1 7 1¼ using matrix inverse method. solve the system of equations: ª1 1 1º « » 16. 1 2 3 Give A = « » find its inverse and hence » ¬149¼ x + y + z = 6, x + 2y + 3z = 14 and x + 4y + 9z = 36. ª1 2 1º 1 verify whether A3 –23A – 40I = 0 where I is the unit matrix of » « » 17. If A = « 3 2 » « 1¼ ¬4 2 order 3. 18. Using matrices A , B and C verify the rule (A.B) C = A (BC) where

ª1 1 0º ª 1 0º 0 11 « » «» A= 01 1» C= «0 1» B=« » «» 1 21 « ¬1 1 1¼ ¬ 1 1¼ ª2 4 0º «» 19. Find the inverse of the matrix « 344 « » » ¬ 2 2 5¼ 20. Given A = 5 14 5 71 8 12 3 2 7 ,B= 3 14 2 34 Find (i) AB + BA (ii) A'B + B'A CU IDOL SELF LEARNING MATERIAL (SLM)

Matrices 93 2 12 21. Given A = 3 4 5 find the inverse of A' 1 13 21 1 22. If A = 1 3 2 show that AA–1 = I 22 1 x+p q r 23. Solve r x+q p [Ans: x = 0 , x = – (p + q + r)] p =0 q x+r 1a a2 bc 1b 24. Prove that b2 ca =0 1c c2 - ab 21 1 25. If A = 1 21 show that 3 – 2 + 9A – 4I = 0 A 6A 11 2

4.14 References References of this unit have been given at the end of the book.  CU IDOL SELF LEARNING MATERIAL (SLM)

94 Business Mathematics and Statistics UNIT 5 DETERMINANTS Structure 5.0 Learning Objectives 5.1 Introduction 5.2 Simultaneous Equations 5.3 The Linear Equations in two Unknowns 5.4 Theorems of Determinants 5.5 Summary 5.6 Key Words/Abbreviations 5.7 Learning Activity

5.8 Unit End Questions (MCQ and Descriptive) 5.9 References 5.0 Learning Objectives After studying this unit, you will be able to:  Explain the meaning and definition of a determinants and note the simple examples to begin with.  Illustrate the method of solving linear equatiions in two unknowns.   Study the solved examples for finding the ‘consistence’.   Elaborate the method of solving a determinant and the 6 different theorems, along the illustrative examples. CU IDOL SELF LEARNING MATERIAL (SLM)

Determinants 95 5.1 Introduction Meaning and Definition of a Determinant The expression ab' – a'b is represented by 7. b a' b' which is called a determinant of the second order. It has two rows a, b and a', b' and two columns a, a' and b, b'. The expression ab' – a'b is called the expansion of the determinant. \\emdash b a' b' a, b, a' , b' are called the elements, ab' and – a'b are called the terms of the expansion. The first ab’ is called the leading term. Example 1: Evaluate =8 59 The expansion is 7 × 9 – 8 × 5 = 63 – 40 = 23 Example 2: Simplify